\(\int \frac {(a+a \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [538]

Optimal result

Integrand size = 41, antiderivative size = 146 \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a (3 A+5 (B+C)) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (A+B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/5*a*A*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/3*a*(A+B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2/5*a*(3*A+5*B+5*C)*(cos(1/2*d
*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/
d+2/3*a*(A+B+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*
x+c)^(1/2)*sec(d*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4159, 4132, 3856, 2719, 4130, 2720} \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 a (A+B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a (3 A+5 (B+C)) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (A+B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + a*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(2*a*(3*A + 5*(B + C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*a*(A + B +
3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a*A*Sin[c + d*x])/(5*d*Sec[c
+ d*x]^(3/2)) + (2*a*(A + B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4159

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int \frac {-\frac {5}{2} a (A+B)-\frac {1}{2} a (3 A+5 (B+C)) \sec (c+d x)-\frac {5}{2} a C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2}{5} \int \frac {-\frac {5}{2} a (A+B)-\frac {5}{2} a C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{5} (a (3 A+5 (B+C))) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} (a (A+B+3 C)) \int \sqrt {\sec (c+d x)} \, dx+\frac {1}{5} \left (a (3 A+5 (B+C)) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = \frac {2 a (3 A+5 (B+C)) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{3} \left (a (A+B+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {2 a (3 A+5 (B+C)) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (A+B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a (A+B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 5.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21 \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a e^{-i d x} \sqrt {\sec (c+d x)} (\cos (d x)+i \sin (d x)) \left (20 (A+B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-4 i (3 A+5 (B+C)) e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) (6 i (3 A+5 (B+C))+10 (A+B) \sin (c+d x)+3 A \sin (2 (c+d x)))\right )}{30 d} \]

[In]

Integrate[((a + a*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(5/2),x]

[Out]

(a*Sqrt[Sec[c + d*x]]*(Cos[d*x] + I*Sin[d*x])*(20*(A + B + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] -
(4*I)*(3*A + 5*(B + C))*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2
*I)*(c + d*x))] + 2*Cos[c + d*x]*((6*I)*(3*A + 5*(B + C)) + 10*(A + B)*Sin[c + d*x] + 3*A*Sin[2*(c + d*x)])))/
(30*d*E^(I*d*x))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(446\) vs. \(2(179)=358\).

Time = 5.32 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.06

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \left (-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (44 A +20 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-16 A -10 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-15 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(447\)
parts	\(-\frac {2 \left (a A +a B \right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}+\frac {2 \left (a B +C a \right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 C a \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 a A \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(663\)

[In]

int((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(-24*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6
+(44*A+20*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-16*A-10*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*A*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*A*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+5*B*(sin(1/2*d*x
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*B*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*C*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*C*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.24 \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-5 i \, \sqrt {2} {\left (A + B + 3 \, C\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (A + B + 3 \, C\right )} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (3 \, A + 5 \, B + 5 \, C\right )} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, A a \cos \left (d x + c\right )^{2} + 5 \, {\left (A + B\right )} a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \]

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

1/15*(-5*I*sqrt(2)*(A + B + 3*C)*a*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(A
+ B + 3*C)*a*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqrt(2)*(3*A + 5*B + 5*C)*a*weier
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt(2)*(3*A + 5*B + 5*C)*a
*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*A*a*cos(d*x + c)^2 +
5*(A + B)*a*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d

Sympy [F]

\[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=a \left (\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int C \sqrt {\sec {\left (c + d x \right )}}\, dx\right ) \]

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2),x)

[Out]

a*(Integral(A/sec(c + d*x)**(5/2), x) + Integral(A/sec(c + d*x)**(3/2), x) + Integral(B/sec(c + d*x)**(3/2), x
) + Integral(B/sqrt(sec(c + d*x)), x) + Integral(C/sqrt(sec(c + d*x)), x) + Integral(C*sqrt(sec(c + d*x)), x))

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)/sec(d*x + c)^(5/2), x)

Giac [F]

\[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)/sec(d*x + c)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int(((a + a/cos(c + d*x))*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2),x)

[Out]

int(((a + a/cos(c + d*x))*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(5/2), x)